I am jammed to problem 11 on page 803 here. Earlier on page 797, it claims that the function changes the least when $\langle \nabla f, \bar{e}\rangle=0$ and it changes the most in the direction of the gradient $\nabla f$.

1 Answer
1

$\DeclareMathOperator{\bR}{\mathbf R}$There does appear to be some confusion. Let $p$ be a point in $\bR^3$ and $f\colon \bR^3 \to \mathbf R^3$ a function. If $u$ is a unit vector in $\bR^3$, then we have the notion of the directional derivative of $f$ at $p$ in the direction of $u$, denoted $D_uf(p)$. Finding the direction of "fastest growth" is finding a $u$ which maximizes this quantity.

If $f$ is differentiable at $p$ then it isn't hard to show that $D_uf(p) = f(p) \cdot u$, where $\nabla f(p)$ is the gradient. Let's make sure that we can compute this: in your example,
\[
\nabla f(x, y, z) = (y^2, 2xy + z^3, 3yz^3).
\]
Using the Schwarz inequality you can conclude that $D_uf(p)$ attains its maximum when $u$ points in the direction of $\nabla f(p)$. The minimum occurs when $u$ points in the opposite direction. If we specialize to $p = (1, 1, 1)$ in the example then $\nabla f(p) = (1, 3, 3)$, and $f$ grows quickest in the direction
\[
u = \frac{1}{\sqrt{19}}(1, 3, 3).
\]

However, "least change" means minimizing $|D_uf(a)|$. This magnitude is actually zero when $\nabla f(p) \cdot u = 0$, i.e. when $u$ lies in the plane (I should assume that $\nabla f(p) \neq 0$) through the origin with normal vector $\nabla f(p)$. So any unit vector in that plane (and there are infinitely many) will do. Continuing the example, I need to find a $u = (u_1, u_2, u_3)$ such that
\[
\begin{pmatrix}1 & 3 & 3\end{pmatrix}\begin{pmatrix}u_1 \\ u_2 \\ u_3\end{pmatrix} = u_1 + 3u_2 + 3u_3 = 0.
\]
You could find a basis for this null space, or just notice that $(3, -1, 0)$ works and normalize this to get
\[
u = \frac{1}{\sqrt{10}}(3, -1, 0).
\]